Answer to Question #217033 in Differential Geometry | Topology for Prathibha Rose

Question #217033
Prove that a countable product of connected space is connected
1
Expert's answer
2021-07-19T17:09:20-0400

Let "\\phi:A\u00d7B\\rightarrow \\{0,1\\}" be a continuous function. It suffices to show that "\\phi" is constant.


Now if we fix "a\\in A" then we get a function "f:B \\rightarrow \\{0,1\\}" defined by "b \\mapsto \\phi(a,b)". Since "B" is connected, it follows that "f" is continuous. So "f" is constant, and hence "\\phi" is continuous on all sets of the form "\\{a\\}\u00d7B".


Similarly we can show that "\\phi" is constant on all sets of the form "A\u00d7\\{b\\}".


Next, fix "(a,b) \\in A \u00d7 B". Also, let "(a',b') \\in A \u00d7 B". Then from the results above, we have "\\phi(a,b)=\\phi(a,b')=\\phi(a',b')."

Hence "A\u00d7B" is connected.



Now suppose "(A_i)" is a countable collection of connected sets for "i\\in \\mathbb{N}". We show that the countable Cartesian product "A_1\u00d7A_2\u00d7..." is also connected

Clearly, "A_1" is connected.

Suppose "A_1\u00d7A_2\u00d7...\u00d7A_k" is connected for some "k\\in \\mathbb{N}".

Let "L=A_1\u00d7A_2\u00d7...\u00d7A_k". Then "L\u00d7A_{k+1}" is connected since the Cartesian product of 2 connected spaces is also connected.

So "A_1\u00d7A_2\u00d7...\u00d7A_k\u00d7A_{k+1}" is connected. Hence our assertion holds by induction.


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